Fake proof that $(5)$ is a prime ideal in $\mathbf{Z}[i]$ Here's a (false) proof that $(5)$ is a prime ideal in $\mathbf{Z}[i]$. I've been trying to spot the mistake, but I can't seem to find it.
Here's how the proof goes: to show that $(5)$ is a prime ideal, we will show that the quotient $\mathbf{Z}[i]/(5)$ is a field. Since we can write $\mathbf{Z}[i] \cong \mathbf{Z}[x]/(x^2 + 1)$, we have
$$(\mathbf{Z}[i])/(5) \cong (\mathbf{Z}[x]/(x^2 + 1))/(5) \cong \mathbf{F}_5[x]/(x^2 + 1) \cong \mathbf{F}_5, $$
where the last isomorphism follows since $x^2 +1$ is reducible in $\mathbf{F}_5$. Since $\mathbf{F}_5$ is a field, this shows that $(5)$ is prime. Where's the mistake?
I'm thinking that the isomorphism $(\mathbf{Z}[x]/(x^2 + 1))/(5) \cong \mathbf{F}_5[x]/(x^2 + 1) $ isn't correct, but I'm fairly sure that when quotienting by two ideals, we can switch the order like this; i.e: quotienting by $(x^2 + 1)$ and then by $(5)$ is the same as doing it in the opposite order. Any corrections would be most welcome. Thanks!
 A: If $K$ is any field, $x^2+1$ being reducible in $K[x]$ does not imply $K[x]/(x^2+1) \simeq K$. (Just note that the left hand side has $K$-dimension $2$, as $\dim_K (K[x]/(f)) = \deg(f)$ in general.) Rather, this is $\simeq K \times K$ (unless $\mathrm{char}(K)=2$, but even then it's not $K$, see below).
Actually, check for yourself that if $p = (x-a_1)(x-a_2)...(x-a_n) \in K[x]$, with distinct $a_i$, then $K[x]/(p) \simeq K \times ... \times K$ ($n$-fold product).
Added: It's asked in a comment whether the above is still true for more general rings $R$ instead of a field $K$. Note that we certainly have $R[x]/(x-a) \simeq R$ (via $x \mapsto a$) for any $a \in R$. But since the proof of $K[x]/\prod(x-a_i) \simeq \prod K[x]/(x-a_i)$ is basically the Chinese Remainder Theorem (a.k.a. Sun-Tsu theorem), that depends on whether its conditions still hold for $R$ instead of $K$.
Well, for any commutative ring $R$ and elements $a,b \in R$, the ideals $(x-a)$ and $(x-b)$ in $R[x]$ are coprime if and only if $a-b \in R^\times$. So if all pairwise differences $a_i - a_j (i \neq j)$ are units in $R$, then the conclusion still holds. (Note how for $R=K$ this reduces to the above condition that all the $a_i$ are different from each other.)
In the particularly easy example of $x^2-1=(x-1)(x+1)$ we see that as soon as $2 \in R^\times$, we get $R[x]/(x^2-1) \simeq R[x]/(x-1) \times R[x]/(x+1) \simeq R \times R$. In particular, this is true for all $R= \mathbb Z/n$ with odd $n$.
However if $2 \notin R^\times$, this can fail; indeed, as alluded to above, already for a field $R=K$ with $\mathrm{char}(K)=2$, the ring $$K[x]/(x^2-1) \simeq K[x]/(x^2+1) \simeq K[x]/(x-1)^2 \; \stackrel{x-1 \;\mapsto \,t}\simeq \; K[t]/(t^2)$$
is not isomorphic to $K \times K$: they share the same underlying additive group, but unlike $K \times K$, the ring above contains an element $x\neq 1$ with $x^2=1$ (or an element $t \neq 0$ with $t^2=0$).
Consequently, we also have $R[x]/(x^2-1) \not \simeq R \times R$ for any ring $R$ such that $R/(2)$ is a field (because such an isomorphism would, modulo $2$, induce an isomorphism in the case of a field which we just excluded). This includes all the $R=\mathbb Z/n$ with even $n$.
As a last remark, I suspect there are rings $R$ with $2 \notin R^\times$ where $R[x]/(x^2-1) \simeq R \times R$ still holds, "accidentally" i.e. in spite of a proof via the Chinese Remainder Theorem failing. I do not have a concrete example for that (in spite of what I wrote here earlier, the candidate I proposed there failed I think), but there are rings $R$ which happen to be $\simeq R \times R$; there are rings $R$ such that $R[x]/(x^2-1) \simeq R$; and I see no reason why there should not be some weird big ring where both these isomorphisms hold, simultaneously, accidentally.
